Question: Simplify the following expression and state the condition under which the simplification is valid. $r = \dfrac{y^3 - 14y^2 + 48y}{-5y^3 - 10y^2 + 400y}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ r = \dfrac {y(y^2 - 14y + 48)} {-5y(y^2 + 2y - 80)} $ $ r = -\dfrac{y}{5y} \cdot \dfrac{y^2 - 14y + 48}{y^2 + 2y - 80} $ Simplify: $ r = - \dfrac{1}{5} \cdot \dfrac{y^2 - 14y + 48}{y^2 + 2y - 80}$ Since we are dividing by $y$ , we must remember that $y \neq 0$ Next factor the numerator and denominator. $ r = - \dfrac{1}{5} \cdot \dfrac{(y - 8)(y - 6)}{(y - 8)(y + 10)}$ Assuming $y \neq 8$ , we can cancel the $y - 8$ $ r = - \dfrac{1}{5} \cdot \dfrac{y - 6}{y + 10}$ Therefore: $ r = \dfrac{ -y + 6 }{ 5(y + 10)}$, $y \neq 8$, $y \neq 0$